Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include "bits/stdc++.h"
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int INF = 1e9;
const ll LINF = 1e18;
inline ll gcd(ll a, ll b) { return b ? gcd(b, a%b) : a; }
inline ll lcm(ll a, ll b) { return a / gcd(a, b)*b; }
template<class S,class T> ostream& operator << (ostream& out,const pair<S,T>& o){ out << "(" << o.first << "," << o.second << ")"; return out; }
template<class T> ostream& operator << (ostream& out,const vector<T> V){ for(int i = 0; i < V.size(); i++){ out << V[i]; if(i!=V.size()-1) out << " ";} return out; }
template<class T> ostream& operator << (ostream& out,const vector<vector<T> > Mat){ for(int i = 0; i < Mat.size(); i++) { if(i != 0) out << endl; out << Mat[i];} return out; }
template<class S,class T> ostream& operator << (ostream& out,const map<S,T> mp){ out << "{ "; for(auto it = mp.begin(); it != mp.end(); it++){ out << it->first << ":" << it->second; if(mp.size()-1 != distance(mp.begin(),it)) out << ", "; } out << " }"; return out; }

/*
 <url:>
 問題文============================================================
 =================================================================
 解説=============================================================
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 */


template<class Type>
Type solve(Type res = Type()){
    int N,M,s,t; cin >> N >> M >> s >> t;
    s--; t--;
    vector<vector<int>> G(N);
    for(int i = 0; i < M;i++){
        int u,v; cin >> u >> v;
        u--; v--;
        G[u].push_back(v);
        G[v].push_back(u);
    }
    vector<int> dists(N,INF),distt(N,INF);
    
    auto dijkstra = [&](int from,vector<int>& dist){
        queue<ll> q;
        dist[from] = 0;
        q.push(from);
        while(q.size()){
            int u = q.front(); q.pop();
            for(auto v:G[u]){
                if(dist[v] > dist[u] + 1){
                    dist[v] = dist[u] + 1;
                    q.push(v);
                }
            }
        }
    };
    
    dijkstra(s,dists);
    dijkstra(t,distt);
    
    map<int,ll> L,R;
    for(int i = 0; i < N;i++){
        if(dists[i] != INF) L[dists[i]]++;
        if(distt[i] != INF) R[distt[i]]++;
    }
    int D = dists[t];
    for(int i = 0; i <= D;i++){
       //if(D-i-2 < 0) continue;
        res += L[i] * R[D-i-2];
    }
    return res;
}
int main(void) {
    cin.tie(0); ios_base::sync_with_stdio(false);
    //test();
    //solve<ll>();
    cout << fixed << setprecision(15) << solve<ll>() << endl;
    return 0;
}